Solving a Rainforest Biologist's Homework Problem

[sweedeljoseph]

Homework Statement

A rain forest biologist is pulling a 50 kg crate of bananas at a constant rate from the ground to the canopy with a rope on a pulley. The distance that the crate has to go is 80 m. When the crate has traveled 40 m, a monkey with a mass of 50 kg jumps on to the top of the crate, and rides on the crate the rest of the way up.
a) How much work does the biologist do on the crate?
b) How much potential energy does the crate have at the top of the tree?

Homework Equations

Wgrav=mg(h_o-h_i)
PEgrav=mgh

The Attempt at a Solution

i remember doing something like this in class before. but there was more than just this. i forgot how to do it though. are you supposed to do it twice then add or something. because another thing is added. like for work you have the normal then when the monkey on it its more. would you just add those together? for the potential energy for the mass you would add both the monkey and the crate? and for the height you would just use the full? you wouldn't use the middle when the monkey jumped on because it wants it at the top right?

thank you!
sweedeljoseph

 

[tiny-tim]

A rain forest biologist is pulling a 50 kg crate of bananas at a constant rate from the ground to the canopy with a rope on a pulley. The distance that the crate has to go is 80 m. When the crate has traveled 40 m, a monkey with a mass of 50 kg jumps on to the top of the crate, and rides on the crate the rest of the way up.
a) How much work does the biologist do on the crate

Hi sweedeljoseph!

Just use the work-energy theorem …

work done = change in energy.

 

[thomate1]

I would approach this problem by breaking it down into smaller parts and using the relevant equations to solve for the answers. Let's start with part (a), which asks for the amount of work the biologist does on the crate.

To solve for work, we can use the equation W = F * d, where W is work, F is force, and d is distance. In this case, the force being applied is the weight of the crate, which is equal to its mass (50 kg) multiplied by the acceleration due to gravity (9.8 m/s^2). So the force is 490 N.

The distance that the crate has to travel is 80 m, but we need to consider the fact that the monkey jumps on at the 40 m mark. This means that the biologist only has to pull the crate for the first 40 m, and then the monkey's weight will help pull it for the remaining 40 m. So the total work done by the biologist is:

W = (490 N * 40 m) + (490 N * 40 m)
= 19,600 J + 19,600 J
= 39,200 J

So the biologist does 39,200 J of work on the crate.

For part (b), we need to calculate the potential energy of the crate at the top of the tree. The equation for potential energy is PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height. In this case, we need to find the potential energy of the crate when it is at the top of the tree, which means we need to use the full height of 80 m.

The mass of the crate is 50 kg, so the potential energy is:

PE = (50 kg * 9.8 m/s^2 * 80 m)
= 39,200 J

Since the monkey's weight is already accounted for in the work calculation, we do not need to add its mass to the potential energy calculation.

Therefore, the crate has 39,200 J of potential energy at the top of the tree.

To summarize, the biologist does 39,200 J of work on the crate, and the crate has 39,200 J of potential energy at the top of the tree. I hope this helps you with your homework problem! Remember to always break down the problem into smaller parts and use the relevant